Question:
The sum $\sum_{k=1}^{20}(1+2+3+\ldots+k)$ is________.
Solution:
Given series can be written as
$\sum_{k=1}^{20} \frac{k(k+1)}{2}=\frac{1}{2} \sum_{k=1}^{20}\left(k^{2}+k\right)$
$=\frac{1}{2}\left[\frac{20(21)(41)}{6}+\frac{20(21)}{2}\right]$
$=\frac{1}{2}\left[\frac{420 \times 41}{6}+\frac{20 \times 21}{2}\right]=\frac{1}{2}[2870+210]=1540$